Class 12th

$P\left(\frac{x\ge 5}{x>2}\right)=\frac{P\left(x\ge 5\cap x>2\right)}{P\left(x>2\right)}=\frac{P\left(x\ge 5\right)}{P\left(x>2\right)}$

$=\frac{1-P\left(x\le 4\right)}{1-P\left(x\le 2\right)}$

$=\frac{1-\left[{\left(\frac{5}{6}\right)}^3*\frac{1}{6}+{\left(\frac{5}{6}\right)}^2*\frac{1}{6}+\left(\frac{5}{6}\right)\frac{1}{6}+\frac{1}{6}\right]}{1-\left[\frac{5}{6}*\frac{1}{6}+\frac{1}{6}\right]}$

$={\left(\frac{5}{6}\right)}^4*{\left(\frac{6}{5}\right)}^2={\left(\frac{5}{6}\right)}^2=\frac{25}{36}$

The number of females who eats Dainik Jagran, The telegram and Dainik Bhaskar:

= (38 + 11 + 21) * 150 = 10500

Required percent $=\frac{10500}{35000}*100=30\mathrm{\%}$

${\overrightarrow{n}}_1=6\widehat{i}+7\widehat{j}+8\widehat{k}{\overrightarrow{n}}_2=3\widehat{i}+5\widehat{j}+7\widehat{k}$

${\overrightarrow{n}}_1*{\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 6\hfill & \hfill 7\hfill & \hfill 8\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 7\hfill \end{array}\right|=9\widehat{i}-18\widehat{j}+9k$              

$\therefore$ the normal to required plane is $\widehat{i}-2\widehat{j}+\widehat{k}$

Equation of plane 1(x + 1) – 2(y – 1) + (z – 3) = 0

x – 2y + z = 0

P (7, -2, 13)

$\therefore PQ=\left|\frac{7+4+13}{\sqrt[]{1+4+1}}\right|=\frac{24}{\sqrt[]{6}}$

${\left(PQ\right)}^2=\frac{24*24}{6}=96$

Required answer:

$=(350*16)+\frac{1}{5}(150*11)=5600+330=5930$

Required number of females:

$=\frac{32}{100}*15000=4800$

$\left|2A\right|=2^3\left|A\right|$

replace A by adj 2A

  $\left|2adj2A\right|=2^3\left|adj2A\right|$           

$=2^3{\left|2A\right|}^2=2^3{\left(2^3\left|A\right|\right)}^2$

$=2^9{\left|A\right|}^2$              

Again replace A by (adj A)

|2adj 2 (adj (adj 2A)| = 2⁹ |adj 2A|⁴

= 2⁹ (|2A|²)⁴

->|A²| = 4

Elemente- gain enthalpy

F-333

Cl-349

Te-190

Po-174

$f\left(x\right)=2x^3-3x^2-12x$  

  $f\text{'}\left(x\right)=6x^2-6x-12$             

= 6 (x – 2) (x + 1)

a = -1, b = 2

$A={\displaystyle \underset{-1}{\overset{0}{\int }}\left(2x^3-3x^2-12x\right)dx-{\displaystyle \underset{0}{\overset{2}{\int }}\left(2x^3-3x^2-12x\right)dx=\frac{57}{2}}}$

->4A = 114

Kindly consider the following figure

Equation of required plane

$\left(x+y+4z-16\right)+\lambda \left(-x+y+z-6\right)=0$ it passes (1, 2, 3)

$\Rightarrow -1+\lambda \left(-2\right)=0$

$\Rightarrow \lambda =-\frac{1}{2}$

$\therefore$ Equation of plane

$\left(1-\lambda \right)x+\left(1+\lambda \right)y+\left(4+\lambda \right)z-16-6\lambda =0$              

$\frac{3}{2}x+\frac{1}{2}y+\frac{7}{2}z-13=0$              

$\Rightarrow 3x+y+7z=26$              

  (4, 2, 2) not satisfying the plane